HDU-1238 Substrings

Substrings

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 6238    Accepted Submission(s): 2767

Problem Description

You are given a number of case-sensitive strings of alphabetic characters, find the largest string X, such that either X, or its inverse can be found as a substring of any of the given strings.

 

Input

The first line of the input file contains a single integer t (1 <= t <= 10), the number of test cases, followed by the input data for each test case. The first line of each test case contains a single integer n (1 <= n <= 100), the number of given strings, followed by n lines, each representing one string of minimum length 1 and maximum length 100. There is no extra white space before and after a string.

 

Output

There should be one line per test case containing the length of the largest string found.

 

Sample Input

2

3

ABCD

BCDFF

BRCD

2

rose

orchid

Sample Output

2

2

 

Author

Asia 2002, Tehran (Iran), Preliminary

 

Recommend

Eddy

 //寻找最长字串并输出其长度
 //  思路：
 //  1、先寻找其中最短字符串str，求子字符串就要先找最短的字符串
 //  2、根据str搜索满足条件的子字符串
 // 3、对str的各子字符串从长到短一次判断是否满足条件，直到找到一个符合条件的子字符串为止
 //strlen：计算字符串长度
 //strstr：在字符串中寻找子字符串,它的用法用的少，但是碰到了，就要熟练。
 //strncpy：复制字符串中的字串，它的用法要掌握。
 //strcpy：复制字符串
 //怎样找最短字符串，也是值得学习的。。
 //搜索满足条件的最长子串.
 #include<stdio.h>
 #include<string.h>
 char a[][];
 int n;
 void cmp(char *str)//字符串反序
 {
     int l,i,g=;
     char b[];
     l=strlen(str);
     for(i=;i<l;i++)
         b[g++]=str[l-i-];
     b[g]='\0';
     strcpy(str,b);
 }
 int search(char *str)//搜索满足条件的最长子串并返回其长度
 {
     int  l1,l2,i,j,f;
     char s[],r[];
     l1=strlen(str);l2=strlen(str);
       while(l1>)//搜索不同长度的子串，从最长的子串开始搜索
       {
           for(i=;i<=l2-l1;i++)
           {
               strncpy(s,str+i,l1);//strncpy的函数用法必须是str+i不能写成str【i】；
               strncpy(r,str+i,l1);
               s[l1]=r[l1]='\0';
               cmp(r);
               f=;
             for(j=;j<n;j++)
                 if(strstr(a[j],s)==NULL && strstr(a[j],r)==NULL)
                 {
                     f=;
                     break;
                 }
                 if(f==)
                     return l1;
           }
           l1--;
       }
       return ;
 }

 int main()
 {
     int i,minstrlen,substrlen,t;//minstrlen记录最短字符串的长度，substrlen返回最长子串长度
       char minstr[];
     scanf("%d",&t);
     while(t--)
     {
         scanf("%d",&n);
         minstrlen=;//先假设字符串最小长度值为100
         for(i=;i<n;i++)
         {
                scanf("%s",a[i]);
              if(strlen(a[i])<minstrlen)//寻找n个字符串中最短的字符串
              {
                  strcpy(minstr,a[i]);
                  minstrlen=strlen(minstr);
              }
         }
         substrlen=search(minstr);
       printf("%d\n",substrlen);
     }
     return ;
 }